Q.Describe any five general methods of preparations of carboxylic acids. OR Giving chemical equations write a brief account of the following:

Five standard routes to RCOOH: oxidise 1° alcohols/aldehydes; oxidise alkylbenzenes; hydrolyse nitriles; hydrolyse esters/amides; carbonate a Grignard reagent. (OR: three named carbonyl reactions — aldol condensation, Cannizzaro reaction, Rosenmund reduction.)

Five general methods of preparing carboxylic acids:

1. Oxidation of primary alcohols or aldehydes: using strong oxidising agents like acidic/alkaline $KMnO_4$ or $K_2Cr_2O_7$:

$RCH_2OH \xrightarrow{[O]} RCHO \xrightarrow{[O]} RCOOH$

2. Oxidation of alkylbenzenes (side chain oxidation): alkyl side chains on benzene, however long, are oxidised right back to a single –COOH group directly on the ring:

$C_6H_5CH_3 \xrightarrow{KMnO_4/K_2Cr_2O_7,\ H^+} C_6H_5COOH$ (benzoic acid)

3. Hydrolysis of nitriles (cyanides): nitriles, made from alkyl halides + $KCN$, are hydrolysed (acid or base catalysed, via the amide) to carboxylic acids:

$RCN + 2H_2O/H^+ \xrightarrow{\Delta} RCOOH + NH_4^+$

4. Hydrolysis of esters or acid amides: acid or base (saponification) hydrolysis of an ester regenerates the carboxylic acid (as its salt, if base-catalysed, then acidified); amides similarly hydrolyse to the acid + ammonia.

5. From Grignard reagents (carbonation): a Grignard reagent reacts with dry ice / CO2 gas to form the magnesium salt of the acid, which is hydrolysed with dilute acid:

$RMgX + CO_2 \longrightarrow RCOOMgX \xrightarrow{H_3O^+} RCOOH$

OR: Aldol condensation, Cannizzaro reaction, Rosenmund reduction

(a) Aldol condensation: aldehydes/ketones possessing at least one α-hydrogen undergo self-addition in the presence of a dilute base (e.g. dil. NaOH) — the enolate of one molecule attacks the carbonyl carbon of another, giving a β-hydroxy aldehyde/ketone ('aldol'). On heating, this readily loses water (dehydration) to give an α,β-unsaturated carbonyl compound:

$2CH_3CHO \xrightarrow{dil.\ NaOH} CH_3CH(OH)CH_2CHO \xrightarrow{\Delta,\ -H_2O} CH_3CH=CHCHO$

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